Question

A particle is thrown upwards from ground. It experiences a constant air resistance force which can produce a retardation of $2m/s^2$. The ratio of time of ascent to the time of descent is:

• A. $1:1$
• B. $\sqrt{\frac{2}{3}}$
• C. $\frac{2}{3}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

Answer: (B)

Let $a$ be the retardation produced by resistive force.

Also let $t_a$ and $t_d$ be the time ascent and descent respectively.

If the particle rises upto a height $h$

Then $h=\frac{1}{2}(g+a)t_a^2$ and $h=\frac{1}{2}(g-a)t_d^2$

$\therefore \frac{t_a}{t_d}=\sqrt{\frac{g-a}{g+a}}$

$=\sqrt{\frac{10-2}{10+2}}=\sqrt{\frac{2}{3}}$